Question

If only $\frac{1}{51}^{th}$ ot the main current is to be passed through a galvanometer then the shunt required is $R_1$ and if only $\frac{1}{11}^{th}$ of the main voltage is to be developed across the galvanometer, then the resistance required is $R_2$. Then $\frac{R_2}{R_1} =$

• A. $\frac{1}{500}$
• B. $\frac{50}{9}$
• C. $\frac{500}{3}$
• D. $500$

Answer 1

Answer: (D)

$500$

Answer 2

According to the question,

If only $\frac{1}{51}$ th of the main current $i$ is to be passed through galvanometer $G$ then the shunt required is main current, $i=51$
$i_{g}=1$
$\therefore R_{1}=\frac{G}{i-i_{9}}$
$\Rightarrow R_{1}=\frac{G}{51-1}=\frac{G}{50} \,\,\,\,\,....(i)$

If only $\frac{1}{11} th$ of the main voltage is developed across the $G$ then the resistance required, $R_{2}$.
$R_{2}=G\left(V_{G}-1\right)$
$R_{2}=G(11-1)=10 G\,\,\,\,\,\,\,\,\dots(ii)$
Now, from Eqn. (i) and (ii), we get
$\frac{R_{2}}{R_{1}}=\frac{10 G}{\frac{G}{50}}=500$
$\therefore \frac{R_{2}}{R_{1}}=500$